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Proof step 8 from proof blastN-len (previous step)

x ≠ [] → (∃ y, a. x = y + a '), # x = n0 + n, ∀ x0, n0, n. # x0 < # x → # x0 = n0 + n → # butlastn(n, x0) = n0 ⊦ x = [], # butlastn(n, x) = n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
# y + 1 = n0 + n, ∀ x0, n0, n. # x0 < #(y + a ') → # x0 = n0 + n → # butlastn(n, x0) = n0 ⊦ # butlastn(n, y + a ') = n0

Rule argument

Additional information

Used simplifier rules: in-17: ⊦ a ∈ b ' ↔ a = b; from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%e-02: ⊦ x + a ' ≠ []; from specification list%len-02: ⊦ # a ' = 0 +1; from specification list%len: ⊦ #(x + y) = # x + # y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%in-15: ⊦ x + a ' = y → a ∈ y; (forward) from specification list